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  4. Ray Optics And Optical Instruments Chapter-Wise Test 3

Ray Optics And Optical Instruments Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A convex lens (\( f = 50 \, \text{cm} \)) and a concave lens (\( f = 25 \, \text{cm} \)) are in contact. What is the effective focal length?

\( f_1 = 50 \, \text{cm} \), \( f_2 = -25 \, \text{cm} \).

\( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{50} + \frac{1}{-25} = \frac{1 - 2}{50} = \frac{-1}{50} \).

\( f = -50 \, \text{cm} \) (diverging system).

-40 cm
-45 cm
-50 cm
-55 cm
3

A light beam converges to a point \( 15 \, \text{cm} \) away. A convex lens of focal length \( 10 \, \text{cm} \) is placed \( 5 \, \text{cm} \) from the convergence point. Where does the beam converge now?

Object distance: \( u = -5 \, \text{cm} \) (virtual object).

\( f = 10 \, \text{cm} \).

\( \frac{1}{v} - \frac{1}{-5} = \frac{1}{10} \Rightarrow \frac{1}{v} + \frac{1}{5} = \frac{1}{10} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{5} = \frac{1 - 2}{10} = \frac{-1}{10} \).

\( v = -10 \, \text{cm} \) (10 cm to the left of the lens).

5 cm
10 cm
15 cm
20 cm
2

A convex mirror of focal length \( 24 \, \text{cm} \) forms an image \( 8 \, \text{cm} \) behind the mirror. What is the object distance?

Focal length: \( f = 24 \, \text{cm} \) (convex mirror).

Image distance: \( v = 8 \, \text{cm} \) (virtual image).

Mirror equation: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \).

\( \frac{1}{8} + \frac{1}{u} = \frac{1}{24} \Rightarrow \frac{1}{u} = \frac{1}{24} - \frac{1}{8} = \frac{1 - 3}{24} = \frac{-2}{24} = \frac{-1}{12} \).

\( u = -12 \, \text{cm} \).

10 cm
12 cm
15 cm
18 cm
1

An object is placed \( 15 \, \text{cm} \) from a convex mirror of focal length \( 30 \, \text{cm} \). What is the magnification?

Focal length: \( f = 30 \, \text{cm} \), \( u = -15 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{-15} = \frac{1}{30} \Rightarrow \frac{1}{v} = \frac{1}{30} + \frac{1}{15} = \frac{1 + 2}{30} = \frac{3}{30} = \frac{1}{10} \).

\( v = 10 \, \text{cm} \).

Magnification: \( m = -\frac{v}{u} = -\frac{10}{-15} = 0.67 \).

0.5
0.6
0.7
0.67
4

A convex lens of focal length \( 12 \, \text{cm} \) forms an image at \( 24 \, \text{cm} \) from the lens. What is the object distance?

Focal length: \( f = 12 \, \text{cm} \).

Image distance: \( v = 24 \, \text{cm} \) (real image).

Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).

\( \frac{1}{24} - \frac{1}{u} = \frac{1}{12} \Rightarrow \frac{1}{u} = \frac{1}{24} - \frac{1}{12} = \frac{1 - 2}{24} = \frac{-1}{24} \).

\( u = -24 \, \text{cm} \).

18 cm
24 cm
30 cm
36 cm
2

A prism of angle \( 60^\circ \) has a minimum deviation of \( 36^\circ \). What is the refractive index?

Refractive index: \( n = \frac{\sin \left( \frac{A + D_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \).

\( A = 60^\circ \), \( D_m = 36^\circ \).

\( n = \frac{\sin \left( \frac{60 + 36}{2} \right)}{\sin \left( \frac{60}{2} \right)} = \frac{\sin 48^\circ}{\sin 30^\circ} \).

\( \sin 48^\circ \approx 0.743 \), \( \sin 30^\circ = 0.5 \).

\( n = \frac{0.743}{0.5} \approx 1.486 \).

1.4
1.49
1.5
1.6
2

A concave lens of focal length \( 25 \, \text{cm} \) forms an image \( 10 \, \text{cm} \) from the lens. What is the object distance?

Focal length: \( f = -25 \, \text{cm} \) (concave lens).

Image distance: \( v = -10 \, \text{cm} \) (virtual image).

Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).

\( \frac{1}{-10} - \frac{1}{u} = \frac{1}{-25} \Rightarrow \frac{1}{u} = \frac{1}{-10} - \frac{1}{-25} = \frac{-5 + 2}{50} = \frac{-3}{50} \).

\( u = -\frac{50}{3} \approx -16.67 \, \text{cm} \).

15 cm
16.7 cm
20 cm
25 cm
2

In a concave lens, what is the effect on the image if the object is moved closer to the lens from a distant position?

For a concave lens, the image is always virtual, erect, and diminished. As the object moves closer, the image size increases (though still smaller than the object), and the image moves closer to the lens, but remains on the same side as the object.

Image becomes real and inverted
Image size increases but remains diminished
Image size decreases further
Image moves to the opposite side
2

A convex mirror of focal length \( 20 \, \text{cm} \) has an object placed \( 40 \, \text{cm} \) from it. What is the image distance?

Focal length: \( f = 20 \, \text{cm} \) (convex mirror).

Object distance: \( u = -40 \, \text{cm} \).

Mirror equation: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \).

\( \frac{1}{v} + \frac{1}{-40} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} + \frac{1}{40} = \frac{2 + 1}{40} = \frac{3}{40} \).

\( v = \frac{40}{3} \approx 13.33 \, \text{cm} \) (virtual image).

10 cm
13.3 cm
15 cm
20 cm
1

A prism with refracting angle \( 60^\circ \) has a minimum deviation of \( 40^\circ \). What is the refractive index of the prism material?

Refractive index: \( n = \frac{\sin \left( \frac{A + D_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \).

\( A = 60^\circ \), \( D_m = 40^\circ \).

\( n = \frac{\sin \left( \frac{60 + 40}{2} \right)}{\sin \left( \frac{60}{2} \right)} = \frac{\sin 50^\circ}{\sin 30^\circ} \).

\( \sin 50^\circ \approx 0.766 \), \( \sin 30^\circ = 0.5 \).

\( n = \frac{0.766}{0.5} \approx 1.53 \).

1.4
1.53
1.6
1.7
2

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